Question

A 8.0 µF capacitor is charged by a 11.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R in kO.

I've been using , but I'm not sure if that is right since I just CAN'T get the right answer...I keep getting 370.7 kO (incorrect).

Answer 1

Concepts and reason

Use the concept of charging of the capacitor to solve this problem.

First calculate the time constant by using the concept of time constant of the capacitor.

Find the resistance of the circuit by using the equation for the voltage across the capacitor when it was charging.

Fundamentals

The expression for the time constant of the RC circuit is as follows:

$\tau = RC$

Here,$\tau$is the time constant, R is the resistance of the circuit, and C is the capacitance.

The expression for the voltage across the capacitor when it was charging is as follows:

${V_c} = {V_s}\left( {1 - {e^{ - \;\frac{t}{\tau }}}} \right)$

Here,${V_c}$is the voltage across the capacitor,${V_s}$is the supply voltage, and t is the elapsed time.

The time constant for the RC circuit is,

$\tau = RC$

Substitute $\left( {8.0\;\mu {\rm{F}}} \right)$for C in the above equation.

$\begin{array}{c}\\\tau = \left( {8.0\;\mu {\rm{F}}\left( {\frac{{1\;{\rm{F}}}}{{{{10}^6}\;\mu {\rm{F}}}}} \right)} \right)R\\\\ = \left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R\\\end{array}$

The voltage across the capacitor when it is charging is,

${V_c} = {V_s}\left( {1 - {e^{ - \;\frac{t}{\tau }}}} \right)$

Substitute$\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R$for$\tau$in the above equation.

$\begin{array}{c}\\{V_c} = {V_s}\left( {1 - {e^{ - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R}}}}} \right)\\\\\frac{{{V_c}}}{{{V_s}}} = 1 - {e^{ - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R}}}}\\\\{e^{ - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R}}}} = 1 - \frac{{{V_c}}}{{{V_s}}}\\\end{array}$

Take ln on both sides as follows:

$\begin{array}{c}\\\ln \left( {{e^{ - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R}}}}} \right) = \ln \left( {1 - \frac{{{V_c}}}{{{V_s}}}} \right)\\\\ - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)R}} = \ln \left( {1 - \frac{{{V_c}}}{{{V_s}}}} \right)\\\end{array}$

Rearrange the above equation for R as follows:

$R = - \;\frac{t}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)\ln \left( {1 - \frac{{{V_c}}}{{{V_s}}}} \right)}}$

Substitute 4.00 V for${V_c}$, 11.0 V for${V_s}$, 3.00 s for t in the above equation.

$\begin{array}{c}\\R = - \;\frac{{3.00\;{\rm{s}}}}{{\left( {8.0 \times {{10}^{ - 6}}\;{\rm{F}}} \right)\ln \left( {1 - \frac{{4.00\;{\rm{V}}}}{{11.0\;{\rm{V}}}}} \right)}}\\\\ = \left( {829.67 \times {{10}^3}\;\Omega } \right)\left( {\frac{{1\;{\rm{k}}\Omega }}{{{{10}^3}\;\Omega }}} \right)\\\\ = 829.67\;{\rm{k}}\Omega \\\end{array}$

Ans:

The resistance of the charging RC circuit is$829.83\;{\rm{k}}\Omega$.